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If $G$ is a finite group which acts transitively on $X$, and if $H$ is a normal subgroup of $G$, show that the orbits of the induced action of $H$ on $X$ all have the same size.

I don't know how to prove this problem. I hope to get your help. Thank you.

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If $Hx$ and $Hy$ are two orbits, then by transitivity there exists $g\in G$ with $gx=y$. Then $z\mapsto gz$ is a map $Hx\to Hy$ because for $hx\in Hx$, we have $ghx=hgx=hy\in Hy$. Its obvious inverse $z\mapsto g^{-1}z$ makes it a bijection.

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    $\begingroup$ Why should we have ghx=hgx? $\endgroup$ – Good boy Oct 2 '16 at 14:00
  • $\begingroup$ Thanks.I'll think about it again by myself. $\endgroup$ – Good boy Oct 2 '16 at 14:10
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    $\begingroup$ There is a little bit of a mistake here regarding "$ghx=hgx$". This is the place where the normality of $H$ becomes important. $\endgroup$ – Keith Kearnes Oct 2 '16 at 21:54

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